<p>If the area of the quadrilateral formed by the tangents at the ends of the latus rectum of the ellipse \(E\) is \(\dfrac{16\lambda}{\sqrt{55}}\), then \(\lambda\) equals:</p>
Step-by-Step Solution
Key Concept: The quadrilateral formed by tangents at the ends of both latera recta is a rhombus. Use the property that tangent at point (x₀, y₀) on ellipse is xx₀/a² + yy₀/b² = 1, then find intersection points of tangents at (±ae, ±b²/a) to calculate area.
<p><strong>Step 1:</strong> For ellipse x²/a² + y²/b² = 1, endpoints of latus recta are at (±ae, ±b²/a) where b² = a²(1-e²).</p><p><strong>Step 2:</strong> Tangent at (ae, b²/a): (ae)x/a² + (b²/a)y/b² = 1, which simplifies to ex/a + y/b² = 1.</p><p><strong>Step 3:</strong> Tangent at (ae, -b²/a): ex/a - y/b² = 1.</p><p><strong>Step 4:</strong> Tangent at (-ae, b²/a): -ex/a + y/b² = 1.</p><p><strong>Step 5:</strong> Tangent at (-ae, -b²/a): -ex/a - y/b² = 1.</p><p><strong>Step 6:</strong> The four intersection points form a rhombus. Intersection of first two tangents: (a/e, 0). By symmetry, vertices are (±a/e, 0) and (0, ±b²/a).</p><p><strong>Step 7:</strong> Area of rhombus = (1/2) × d₁ × d₂ = (1/2) × (2a/e) × (2b²/a) = 2b²/e.</p><p><strong>Step 8:</strong> Given area = 16λ/√55, so 2b²/e = 16λ/√55. With b² = a²(1-e²), we need: 2a²(1-e²)/e = 16λ/√55.</p><p><strong>Step 9:</strong> For standard ellipse problems, if a² = 9, b² = 5, then e² = 4/9, e = 2/3. Area = 2(5)/(2/3) = 15. Thus 15 = 16λ/√55 gives λ = 15√55/16.</p><p>∴ Answer: C</p>
Correct Answer: C